Codeforces Round #748 (Div. 3)

比赛链接

Codeforces Round #748 (Div. 3)

D2. Half of Same

Polycarp has an array of $n\left(n\right.$ is even) integers $a_{1}, a_{2}, \ldots, a_{n}$. Polycarp conceived of a positive integer $k$. After that, Polycarp began performing the following operations on the array: take an index $i(1 \leq i \leq n)$ and reduce the number $a_{i}$ by $k$.

After Polycarp performed some (possibly zero) number of such operations, it turned out that at least half of the numbers in the array became the same. Find the maximum $k$ at which such a situation is possible, or print $-1$ if such a number can be arbitrarily large.

Input

The first line contains one integer $t(1 \leq t \leq 10)$ - the number of test cases. Then $t$ test cases follow.
Each test case consists of two lines. The first line contains an even integer $n(4 \leq n \leq 40)$ ( $n$ is even). The second line contains $n$ integers $a_{1}, a_{2}, \ldots a_{n}\left(-10^{6} \leq a_{i} \leq 10^{6}\right)$.
It is guaranteed that the sum of all $n$ specified in the given test cases does not exceed $100$ .

Output

For each test case output on a separate line an integer $k(k \geq 1)$ - the maximum possible number that Polycarp used in operations on the array, or $-1$, if such a number can be arbitrarily large.

Example

input

4
6
48 13 22 -15 16 35
8
-1 0 1 -1 0 1 -1 0
4
100 -1000 -1000 -1000
4
1 1 1 1

output

13
2
-1
-1

解题思路

思维

显然，对于选出来的任意两个数 $a,b$，有 $a\%k=b\%k$，即 $(a-b)\%k=0$，可以枚举选出来的最小值，将所有数减去这个最小值，这些数对 $k$ 取模的模数都为 $0$，即枚举所有差值的因子，如果其出现的次数大于等于 $n/2$，则该因子满足条件，此时更新答案，注意，如果出现其他数和最小值相等，则该数必选，因此时差值为 $0$，肯定满足条件

• 时间复杂度：$O(n^2\times max(\{a_i\}))$

代码

// Problem: D2. Half of Same
// Contest: Codeforces - Codeforces Round #748 (Div. 3)
// URL: https://codeforces.com/contest/1593/problem/D2
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// %%%Skyqwq
#include <bits/stdc++.h>
 
//#define int long long
#define help {cin.tie(NULL); cout.tie(NULL);}
#define pb push_back
#define fi first
#define se second
#define mkp make_pair
using namespace std;
 
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
 
template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }
 
template <typename T> void inline read(T &x) {
    int f = 1; x = 0; char s = getchar();
    while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
}

const int N=45;
int t,n,a[N];
vector<int> Div(int n)
{
	vector<int> res;
	for(int i=1;i<=n/i;i++)
		if(n%i==0)
		{
			res.pb(i);
			if(i*i!=n)
				res.pb(n/i);
		}
	return res;
}
int main()
{
	help;
    for(cin>>t;t;t--)
    {
    	cin>>n;
    	for(int i=1;i<=n;i++)cin>>a[i];
    	sort(a+1,a+1+n);
    	bool f=false;
    	for(int i=1;i+n/2-1<=n;i++)
    		if(a[i]==a[i+n/2-1])
    		{
    			f=true;
    			break;
    		}
    	if(f)
    	{
    		puts("-1");
    		continue;
    	}
    	int res=0;
    	for(int i=1;i<=n;i++)
    	{
    		vector<int> b;
    		int tt=0;
    		for(int j=1;j<=n;j++)
    			if(a[j]>a[i])b.pb(a[j]-a[i]);
    			else if(a[i]==a[j])tt++;
    		unordered_map<int,int> mp;
    		for(int i:b)
    			for(int j:Div(i))
    				mp[j]++;
    		for(auto t:mp)
    			if(t.se>=n/2-tt)res=max(res,t.fi);
    	}
    	cout<<res<<'\n';
    }
    return 0;
}

__EOF__

本文作者：acwing_zyy
本文链接：https://www.cnblogs.com/zyyun/p/16276209.html
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